Qualitative analysis

Hand a chemist an unknown white powder and the first question is never "how much" but "what is even in it." Before any quantity can be measured, the elements present in an organic compound must be confirmed one at a time. Carbon and hydrogen are found by burning the compound with copper(II) oxide and catching the products it releases. Nitrogen, sulphur, halogens, and phosphorus are found by a very different route: fusing the compound with sodium metal to break its covalent bonds and force each heteroatom into a simple, water-soluble ionic form. That single extract, prepared once, then answers four separate yes-or-no questions through four short colour tests. This article walks through both routes, the reasoning behind sodium fusion, and the exact observations that count as a positive result.

The core insight

Carbon and hydrogen are detected by oxidising the compound with copper(II) oxide, while nitrogen, sulphur, halogens, and phosphorus are detected via Lassaigne's sodium fusion extract, in which each heteroatom is converted to a simple sodium salt and then identified by a characteristic colour or precipitate test.

The core idea

Two families of tests, two different chemistries. Detecting carbon and hydrogen relies on oxidation; detecting the other four elements relies on reductive fusion with sodium. Mixing these up — treating them as one procedure — is the single most common conceptual error at this level.

Carbon and hydrogen: oxidation with CuO. The compound is heated with dry copper(II) oxide. Carbon is oxidised to carbon dioxide and hydrogen to water, while $\text{CuO}$ itself is reduced to copper:

\text{C} + 2\text{CuO} \rightarrow \text{CO}_2 + 2\text{Cu}

2\text{H} + \text{CuO} \rightarrow \text{H}_2\text{O} + \text{Cu}

where: $\text{C}$ and $\text{H}$ are the carbon and hydrogen bound in the organic compound, and $\text{CuO}$ is the oxidising agent, itself reduced to metallic copper. $\text{CO}_2$ is confirmed downstream by turning lime water milky, $\text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3\downarrow + \text{H}_2\text{O}$ ; $\text{H}_2\text{O}$ is confirmed by condensing on a cool part of the tube and turning white anhydrous copper sulphate blue.

Nitrogen, sulphur, halogens, phosphorus: Lassaigne's test. These four elements are covalently bound and give no simple ionic test in the intact molecule. Fusing the compound with freshly cut sodium metal at red heat breaks the C–N, C–S, and C–X bonds and forces each heteroatom to combine with sodium (and, for nitrogen, with carbon already present in the molecule), producing water-soluble ionic salts:

\text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN}, \qquad 2\text{Na} + \text{S} \rightarrow \text{Na}_2\text{S}, \qquad \text{Na} + \text{X} \rightarrow \text{NaX}

where: $\text{X}$ stands for any halogen present. Plunging the red-hot fusion tube into distilled water, boiling, and filtering gives the clear sodium fusion extract (Lassaigne's extract) — a single solution now carrying $\text{CN}^-$ , $\text{S}^{2-}$ , and/or $\text{X}^-$ ions, each testable separately. If both nitrogen and sulphur are present together, sodium thiocyanate ( $\text{NaSCN}$ ) can form instead of separate cyanide and sulphide ions, which is why an excess of sodium is always used during fusion — it keeps enough free $\text{CN}^-$ and $\text{S}^{2-}$ available for both tests to stay reliable.

Figure

Flow diagram from an organic compound through sodium fusion to a single extract that branches into four separate confirmation tests

Why it works

Sodium is used, rather than any other reagent, for a specific chemical reason: covalent bonds like C–N, C–S, and C–X do not ionise on their own, so no direct precipitation or colour test can detect them while they remain part of the organic framework. Fusing with reactive, electropositive sodium metal supplies the energy and the reducing power to snap those covalent bonds and re-form the heteroatoms as simple, freely dissociating sodium salts — $\text{NaCN}$ , $\text{Na}_2\text{S}$ , $\text{NaX}$ — which behave exactly like the ionic compounds studied in inorganic qualitative analysis. This is why the extract, and not the original compound, is what actually gets tested in each of the four downstream steps. Each confirmatory test is then chosen because it targets the specific ion produced, and nothing else:

Nitrogen: boiling the extract with freshly prepared iron(II) sulphate, then acidifying with dilute sulphuric acid, converts any $\text{CN}^-$ present into a complex cyanide that combines with partially air-oxidised $\text{Fe}^{3+}$ to give insoluble ferric ferrocyanide — the deep Prussian blue colour or precipitate.
Sulphur: the extract with freshly prepared sodium nitroprusside gives a violet/purple colour if $\text{S}^{2-}$ is present (the reaction is specific to the sulphide ion); alternatively, acidifying with acetic acid and adding lead acetate gives a black precipitate of $\text{PbS}$ .
Halogens: the extract is first acidified with dilute nitric acid — this step destroys interfering $\text{CN}^-$ and $\text{S}^{2-}$ ions, which would otherwise also precipitate with silver — before silver nitrate is added. The colour and solubility of the silver halide formed identifies which halogen is present.
Phosphorus: the compound is oxidised (for example, by fusion with sodium peroxide) to convert bound phosphorus to phosphate, which is then boiled with concentrated nitric acid and treated with ammonium molybdate to give a yellow precipitate of ammonium phosphomolybdate.

Worked example

Question: A Lassaigne extract, when acidified with dilute nitric acid and treated with silver nitrate solution, gives a pale yellow precipitate that is only partially soluble in ammonium hydroxide. The same compound is also known to contain nitrogen. Identify the halogen present, and explain both the role of the nitric acid step and why the ammonium hydroxide solubility test matters.

Step 1. Recall why nitric acid is added first. Because the compound contains nitrogen, the extract also carries $\text{CN}^-$ ions, which would otherwise react with $\text{Ag}^+$ to give a misleading precipitate of their own. Boiling with dilute nitric acid decomposes cyanide (and any sulphide) ions before silver nitrate is added, so only the halide ion reacts cleanly.

Step 2. Interpret the silver nitrate observation. A precipitate that is pale yellow and only partially soluble in ammonium hydroxide is the diagnostic signature of silver bromide, $\text{AgBr}$ — distinct from the fully soluble white $\text{AgCl}$ and the insoluble yellow $\text{AgI}$ .

Step 3. State the conclusion. The compound contains bromine. The nitric acid pre-treatment was essential precisely because nitrogen was also present; without it, cyanide interference could have produced an unreliable precipitate pattern, and the ammonium hydroxide solubility test is what distinguishes the three halides once the interfering ions are gone.

Answer: the compound contains bromine, confirmed by a pale yellow, partially ammonium-hydroxide-soluble silver halide precipitate obtained only after destroying cyanide interference with dilute nitric acid.

Competitive stretch

If the same extract also gave a violet colour with sodium nitroprusside, the compound would contain sulphur as well as nitrogen and bromine — three independent positive tests on one extract, each targeting a different ion with no cross-interference once the nitric acid step has been performed.

Common traps

Treating the CuO oxidation test (for C and H) and the Lassaigne sodium fusion test (for N, S, halogens, P) as one procedure — they use opposite chemistry, oxidation versus reductive fusion, and detect entirely different elements.
Skipping the dilute nitric acid step before the halogen test when nitrogen or sulphur is also present, which lets cyanide or sulphide ions contaminate the silver nitrate result.
Mixing up the three silver halide signatures: $\text{AgCl}$ is white and fully soluble in $\text{NH}_4\text{OH}$ , $\text{AgBr}$ is pale yellow and only partially soluble, $\text{AgI}$ is yellow and insoluble.
Reading a faint or negative Prussian blue test as definite proof of "no nitrogen" — insufficient sodium during fusion, especially with sulphur also present, can form $\text{NaSCN}$ instead of separate $\text{NaCN}$ , so a weak result is only reliable if excess sodium was used at the fusion step.

Ask the tutor: Want to walk through the full Lassaigne test sequence on a sample compound, step by step, together?

Why is sodium metal specifically used for the fusion, and not some other reactive metal?

Sodium is reactive and electropositive enough to break the covalent C–N, C–S, and C–X bonds at fusion temperature and combine directly with the freed heteroatoms, forming simple, water-soluble ionic salts. It is also cheap, easy to cut into small fresh pieces, and its combustion products are easy to handle safely compared with more exotic alternatives. The freshly cut requirement matters because sodium develops an oxide coating in air that would blunt its reactivity during fusion.

Why does the copper sulphate turn blue in the carbon and hydrogen test, and what does that actually prove?

Anhydrous copper sulphate is white because it has no water of crystallisation bound to it. When water vapour, produced by the oxidation of hydrogen in the compound, condenses on the cooler part of the test tube where the anhydrous salt sits, it rehydrates to blue hydrated copper sulphate. The colour change is a simple, unambiguous test for the presence of water, which in turn confirms that hydrogen was present in the original organic compound.

If a compound tests negative for both nitrogen and sulphur but positive for chlorine, is the fusion extract still trustworthy?

Yes, provided excess sodium was used during fusion. The complication of thiocyanate formation (NaSCN instead of separate NaCN and Na2S) only arises when both nitrogen and sulphur are present together and sodium is insufficient; it does not affect the halogen test, which is independent of that interference. A clean negative for N and S alongside a positive halogen result is a perfectly reliable outcome as long as the fusion itself was carried out correctly with fresh, excess sodium.